We have a group of 15 people and want to find out how many possibilities there are to choose a subgroup of 5 people. This is pretty easy, we can just calculate $15 \choose 5$.
But now suppose we have a constraint where two people don't want to be in the subgroup together. My first guess would have been to calculate how many possibilities there are that these two are in the same subgroup (which would be $15 \choose 3$, right?) and then subtract that, so my solution to this problem would look like this:
${15 \choose 5} - {15 \choose 3} = 3003 - 455 = 2548$
Is this the correct way to approach this problem? If now, what would be the right answer?
There are three possible types of subgroups:
a) None of the two persons are in the subgroup
b) Only the first one is in the subgroup
c) Only the second one is in the subgroup
For a) is the same to choose $5$ people out of $13$, i.e. ${13}\choose{5}$
For b) is the same to choose $4$ people out of $13$ (because you have already one and you don't want the other), i.e. ${13}\choose {4}$
For c) is the same as b)
Thus the solution is ${13}\choose {5}$ +$2 $${13}\choose{4}$ $=1287+2 \cdot 715=1287+1430=2717$